Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any one of them.
Two trees are duplicate if they have the same structure with same node values.
Example 1:
1
/ \
2 3
/ / \
4 2 4
/
4
The following are two duplicate subtrees:
2
/
4
and
4
Therefore, you need to return above trees' root in the form of a list.
这道题考的是DFS+序列化二叉树
我们将每一个节点的左子节点的值和右结点的值都存储下来，组成一个字符串，作为索引，将对应节点保存到map里。
如果一样的字符串已经出现过一次了，我们就把他的root保存在要返回的list中：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution { Map<String, Integer> map;
List<TreeNode> list; public List<TreeNode> findDuplicateSubtrees(TreeNode root) {
this.map = new HashMap<>();
this.list = new ArrayList<>();
collect(root);
return list;
}
private String collect (TreeNode node ){
if(node == null){return "#";}
String serial = node.val +","+collect(node.left) +","+collect(node.right);
map.put(serial, map.getOrDefault(serial,0)+1);
if(map.get(serial) ==2 ){
list.add(node);
}
return serial;
}
}